如何在Javascript数组中合并/推送重复值[关闭]

Posted 2023-02-22

【中文标题】如何在Javascript数组中合并/推送重复值[关闭]

【英文标题】：How to merge/push duplicate value in Javascript Array [closed]

【发布时间】：2018-07-01 23:20:11

【问题描述】：

我有一个这样的数组对象：

[
  "item_id":1,"name":"DOTA 2 Backpack","image":"XXX","qty":1,"original_price":1450000,"total_price":1450000, 
  "item_id":2,"name":"Mobile Legend Backpack","image":"XXX","qty":1,"original_price":1350000,"total_price":1350000
]

当我添加这个具有相同值的数组时，会是这样的：

[
  "item_id":1,"name":"DOTA 2 Backpack","image":"XXX","qty":1,"original_price":1450000,"total_price":1450000,
  "item_id":2,"name":"Mobile Legend Backpack","image":"XXX","qty":1,"original_price":1350000,"total_price":1350000,
  "item_id":2,"name":"Mobile Legend Backpack","image":"XXX","qty":1,"original_price":1350000,"total_price":1350000
]

我的目标想要相同的值会这样合并：

    [
  "item_id":1,"name":"DOTA 2 Backpack","image":"XXX","qty":1,"original_price":1450000,"total_price":1450000,
  "item_id":2,"name":"Mobile Legend Backpack","image":"XXX","qty":2,"original_price":1350000,"total_price":2700000
]

我很困惑并坚持下去。请有人帮助我。

【问题讨论】：

你在使用 lodash 吗？ @Axax

不，顺便说一句，我是 javascript 的新手。

不，Item_id 不同。示例：Dota 2 -> item_id : 1 Mobile Legend -> item_id : 2

不，X 是假的。

你的“合并”代码是什么？

【参考方案1】：

您可以使用Array.prototype.reduce().. 在代码示例中，我正在与元素的item_id进行比较。

代码：

const array = ["item_id": 1,"name":"DOTA 2 Backpack","image":"XXX","qty":1,"original_price":1450000,"total_price":1450000, "item_id":2,"name":"Mobile Legend Backpack","image":"XXX","qty":1,"original_price":1350000,"total_price":1350000];
const newVal = "item_id": 2,"name":"Mobile Legend Backpack","image":"XXX","qty":1,"original_price":1350000,"total_price":1350000;

const addElement = (arr, val) => 
  return arr.reduce((a, c, i) => 
    if (c['item_id'] === val['item_id']) 
      a.found = true;
      c.qty += val.qty;
    
    
    a.result.push(c);
    
    return !a.found && i + 1 === arr.length
      ? [].concat(arr.result, val)
      : a;
  , result: [], found: false);
;


console.log(addElement(array, newVal).result);

.as-console-wrapper  max-height: 100% !important; top: 0; 

【讨论】：

非常感谢。解决了！ :)

【参考方案2】：

看看这个。或许能帮到你。

您可以使用Array Iteration methods

var array1 = [
  "item_id":X,"name":"DOTA 2 Backpack","image":"XXX","qty":1,"original_price":1450000,"total_price":1450000, 
  "item_id":X,"name":"Mobile Legend Backpack","image":"XXX","qty":1,"original_price":1350000,"total_price":1350000
];

var array2 = [
  "item_id":X,"name":"DOTA 2 Backpack","image":"XXX","qty":1,"original_price":1450000,"total_price":1450000,
  "item_id":X,"name":"Mobile Legend Backpack","image":"XXX","qty":1,"original_price":1350000,"total_price":1350000,
  "item_id":X,"name":"Mobile Legend Backpack","image":"XXX","qty":1,"original_price":1350000,"total_price":1350000
]

var props = ['name', 'item_id'];

var result = array1.filter(function(o1)

    return !array2.some(function(o2)
        return o1.item_id=== o2.item_id;         
    );
).map(function(o)

    return props.reduce(function(newo, name)
        newo[name] = o[name];
        return newo;
    , );
);

输出：

[
  "item_id":X,"name":"DOTA 2 Backpack","image":"XXX","qty":1,"original_price":1450000,"total_price":1450000,
  "item_id":X,"name":"Mobile Legend Backpack","image":"XXX","qty":1,"original_price":1350000,"total_price":1350000

]

希望对您有所帮助..！

【讨论】：

你读过这个问题吗？这不能回答问题，也不能解决 OP 问题。

非常感谢。解决了！ :)

@Axaxaxaxax 如果解决了。标记一下。谢谢

【参考方案3】：

如果您想将新的object 推送到array，如果它不存在并更新它，如果它存在：

您需要先使用

Array#some()

method 检查它是否存在。 然后使用found 索引更新其qty 和total_price。 或者如果它不存在，只需推送它。

这应该是你的代码：

var found = -1;
if (arr.some(function(el, i) 
    if (el.name === obj.name)
      found = i;
    return el.name === obj.name;
  )) 
  arr[found]["qty"] = arr[found]["qty"] + obj["qty"];
  arr[found]["total_price"] = arr[found]["total_price"] + obj["total_price"];
 else 
  arr.push(obj);

演示：

var arr = [
    "item_id": 'X',
    "name": "DOTA 2 Backpack",
    "image": "XXX",
    "qty": 1,
    "original_price": 1450000,
    "total_price": 1450000
  ,
  
    "item_id": 'X',
    "name": "Mobile Legend Backpack",
    "image": "XXX",
    "qty": 1,
    "original_price": 1350000,
    "total_price": 1350000
  
];

var obj = 
  "item_id": 'X',
  "name": "Mobile Legend Backpack",
  "image": "XXX",
  "qty": 1,
  "original_price": 1350000,
  "total_price": 1350000
;
var found = -1;
if (arr.some(function(el, i) 
    if (el.name === obj.name)
      found = i;
    return el.name === obj.name;
  )) 
  arr[found]["qty"] = arr[found]["qty"] + obj["qty"];
  arr[found]["total_price"] = arr[found]["total_price"] + obj["total_price"];
 else 
  arr.push(obj);


console.log(arr);

【讨论】：

非常感谢。解决了！ :)

@Axaxaxaxax 欢迎您，很高兴它有帮助。如果对您有帮助，您可以接受。

【参考方案4】：

if (items.indexOf(toAdd) != -1)
   items.push(toAdd)

或者你可以使用 jQuery grep，它返回一个包含请求值的数组列表

if  (($.grep(items, function(e) return e.id == toAdd.id; )) == 0)
   items.push(toAdd)

【讨论】：

你知道我的问题吗？